First slide

Introduction to integration

Question

$\int \frac{s i n 2 x}{\sin 5 x \sin 3 x} d x =$

Easy

A

$\frac{1}{3} \log \sin 3 x - \log \sin 5 x + c$
B

$\frac{1}{3} \log \sin 3 x + \frac{1}{5} \log \sin 5 x + c$
C

$\frac{1}{3} \log \sin 3 x - \frac{1}{5} \log \sin 5 x + c$
D

$3 \log \sin 3 x - 5 \log \sin 5 x + c$

Solution

$∫ \frac{s i n 2 x}{s i n 5 x s i n 3 x} d x = \int \frac{s i n (5 x - 3 x)}{s i n 5 x s i n 3 x} d x = \int \frac{s i n 5 x c o s 3 x - c o s 5 x s i n 3 x}{s i n 5 x s i n 3 x} d x = \int (c o t 3 x - c o t 5 x) d x = \frac{1}{3} l o g s i n 3 x - \frac{1}{5} l o g s i n 5 x + c$

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